I recommend the following approach, in this case. Let $y=\sqrt{x-1}$. Now you need only solve the equation $|y-2|+|y-3|=1$--which should have a closed interval's worth of solutions--then given any of the solutions, say $\alpha$, solve the equation $\sqrt{x-1}=\alpha$. In the end, you will obtain a closed interval's worth of solutions.

It is important to note that this approach will not always work! If the equation you'd started with had been $$\left|\sqrt{x+1}-2\right|+\left|\sqrt{x}-3\right|=1,$$ we could not have made the substitution as above.

$\begingroup$Why will it not always work?$\endgroup$
– mghJun 5 '12 at 0:58

$\begingroup$The problem lies in the fact that $\sqrt{x+1}$ and $\sqrt{x}$ are not linearly related. If we were to try the substitution $y=\sqrt{x+1}$, then we'd need $\sqrt{x}=\sqrt{y^2-1}$, and that's no good. We face a similar problem if we try the substitution $y=\sqrt{x }$.$\endgroup$
– Cameron BuieJun 5 '12 at 1:12